Question

A 0.020-kg bullet is shot horizontally and collides with a \(2.00-\mathrm{kg}\) block of wood. The bullet embeds in the block and the block slides along a horizontal surface for \(1.50 \mathrm{m} .\) If the coefficient of kinetic friction between the block and surface is \(0.400,\) what was the original speed of the bullet?

Short answer

Answer: The original speed of the bullet was approximately 167.76 m/s.

Step-by-step solution

Conservation of Momentum

Before the collision, we have the bullet moving with an initial velocity (v_bullet_initial) and the wooden block at rest. After the collision, both bullet and block move together with a combined velocity (v_combined). We can use the conservation of momentum to find the combined velocity. Momentum_before = Momentum_after (m_bullet * v_bullet_initial) + (m_block * 0) = (m_bullet + m_block) * v_combined Now we can solve the equation for v_combined: v_combined = (m_bullet * v_bullet_initial) / (m_bullet + m_block)

Work-Energy Principle

Now that we have v_combined, we can use the work-energy principle to relate the work done by friction to the change in kinetic energy: Work_done = Change_in_Kinetic_Energy (Force_friction * distance) = (1/2 * (m_bullet + m_block) * v_combined^2) - (1/2 * (m_bullet + m_block) * 0^2) The force of friction can be found using the equation: Force_friction = coefficient_of_friction * Normal_Force Since the block is on a horizontal surface, the normal force is equal to the gravitational force acting on the block: Normal_Force = (m_bullet + m_block) * g Substituting Normal_Force in the Work_done equation, we get: Work_done = -coefficient_of_friction * (m_bullet + m_block) * g * distance We used a negative sign before the coefficient_of_friction because the friction force opposes the motion of the block. Now, plug in the given values and solve for v_bullet_initial: -0.400 * (0.020 + 2.00) * 9.81 * 1.50 = (1/2 * (0.020 + 2.00) * v_combined^2)

Solve for the initial bullet speed

Now we have two equations: v_combined = (0.020 * v_bullet_initial) / (0.020 + 2.00) and -0.400 * (0.020 + 2.00) * 9.81 * 1.50 = (1/2 * (0.020 + 2.00) * v_combined^2) First, solve the second equation for v_combined: v_combined^2 = (-0.400 * (0.020 + 2.00) * 9.81 * 1.50) / (0.5 * (0.020 + 2.00)) Now substitute the first equation into the second equation to find the initial speed of the bullet: v_bullet_initial = ((0.020+2.00) * (-0.400 * (0.020 + 2.00) * 9.81 * 1.50) / (0.5 * (0.020 + 2.00))) / 0.020 Performing the calculation, we obtain: v_bullet_initial ≈ 167.76 m/s So, the initial speed of the bullet was approximately 167.76 m/s.